2.5 The Mole Concept

We can argue that modern chemical science began when scientists started exploring the quantitative as well as the qualitative aspects of chemistry. For example, Dalton’s atomic theory was an attempt to explain the results of measurements that allowed him to calculate the relative masses of elements combined in various compounds. Understanding the relationship between the masses of atoms and the chemical formulas of compounds allows us to quantitatively describe the composition of substances.

The Mole

The identity of a substance is defined not only by the types of atoms or ions it contains, but by the quantity of each type of atom or ion. For example, water, $\mathrm{H}{\phantom{A}}_2\mathrm{O}$, and hydrogen peroxide, $\mathrm{H}{\phantom{A}}_2\mathrm{O}{\phantom{A}}_2$, are alike in that their respective molecules are composed of hydrogen and oxygen atoms. However, because a hydrogen peroxide molecule contains two oxygen atoms, as opposed to the water molecule, which has only one, the two substances exhibit very different properties. Today, we possess sophisticated instruments that allow the direct measurement of these defining microscopic traits; however, the same traits were originally derived from the measurement of macroscopic properties (the masses and volumes of bulk quantities of matter) using relatively simple tools (balances and volumetric glassware). This experimental approach required the introduction of a new unit for amount of substances, the mole, which remains indispensable in modern chemical science.

The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter. One Latin connotation for the word “mole” is “large mass” or “bulk,” which is consistent with its use as the name for this unit. The mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental property, number of atoms, molecules, and so forth. A mole of a substance is that amount in which there are 6.02214179 × 10²³ discrete entities (atoms or molecules). This larger number is a fundamental constant known as Avogadro’s number (N_A) or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly reported with an explicit unit of “per mole,” a conveniently rounded version being 6.022 × 10²³ mol.

Consistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol) (see Figure 2.5.1).

The atomic mass of any substance is numerically equivalent to its atomic or formula weight in amu. Per the amu definition, a single ${\phantom{A}}_{\phantom{}}^{\phantom{12}}{\phantom{A}}_{\phantom{2}}^{\phantom{2}12}\mathrm{C}$ atom weighs 12 amu (its atomic mass is 12 amu). A mole of ${\phantom{A}}_{\phantom{}}^{\phantom{12}}{\phantom{A}}_{\phantom{2}}^{\phantom{2}12}\mathrm{C}$ weighs 12 g (its molar mass is 12 g/mol). This relationship holds for all elements, since their atomic masses are measured relative to that of the amu-reference substance, ${\phantom{A}}_{\phantom{}}^{\phantom{12}}{\phantom{A}}_{\phantom{2}}^{\phantom{2}12}\mathrm{C}$. Extending this principle, the molar mass of a compound in grams is likewise numerically equivalent to its formula mass in amu (Figure 2.5.2).

While atomic mass and molar mass are numerically equivalent, keep in mind that they are vastly different in terms of scale, as represented by the vast difference in the magnitudes of their respective units (amu versus g). To appreciate the enormity of the mole, consider a small drop of water weighing about 0.03 g (see Figure 2.5.3). Although this represents just a tiny fraction of 1 mole of water (~18 g), it contains more water molecules than can be clearly imagined. If the molecules were distributed equally among the roughly seven billion people on earth, each person would receive more than 100 billion molecules.

Worked out examples

The relationships between molar mass, the mole, and Avogadro’s number can be applied to compute various quantities that describe the composition of substances and compounds, as demonstrated in the next several example problems.

Example 2.5.1: Deriving Moles from Grams for an Element

According to nutritional guidelines from the US Department of Agriculture, the estimated average requirement for dietary potassium is 4.7 g. What is the estimated average requirement of potassium in moles?

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The mass of $\mathrm{K}$ is provided, and the corresponding amount of $\mathrm{K}$ in moles is requested. Referring to the periodic table, the atomic mass of $\mathrm{K}$ is 39.10 amu, and so its molar mass is 39.10 g/mol. The given mass of $\mathrm{K}$ (4.7 g) is a bit more than one-tenth the molar mass (39.10 g), so a reasonable “ballpark” estimate of the number of moles would be slightly greater than 0.1 mol.

The molar amount of a substance may be calculated by dividing its mass (g) by its molar mass (g/mol):

The factor-label method supports this mathematical approach since the unit “g” cancels and the answer has units of “mol:”

$4.7\cancel{\text{g}}\mathrm{K}\left({\displaystyle \frac{\text{mol }\mathrm{K}}{39.10\cancel{\text{g}}}}\right)=0.12\text{mol }\mathrm{K}$

The calculated magnitude (0.12 mol $\mathrm{K}$) is consistent with our ballpark expectation, since it is a bit greater than 0.1 mol.

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Example 2.5.2: Deriving Grams from Moles for an Element

A liter of air contains 9.2 × 10^-4 mol argon. What is the mass of $\mathrm{Ar}$ in a liter of air?

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The molar amount of $\mathrm{Ar}$ is provided and must be used to derive the corresponding mass in grams. Since the amount of $\mathrm{Ar}$ is less than 1 mole, the mass will be less than the mass of 1 mole of $\mathrm{Ar}$, approximately 40 g. The molar amount in question is approximately one-one thousandth 10^-3 of a mole, and so the corresponding mass should be roughly one-one thousandth of the molar mass 0.04 g:

In this case, logic dictates (and the factor-label method supports) multiplying the provided amount (mol) by the molar mass (g/mol):

$9.2\times {10}^{-4}\cancel{\text{mol}}\text{Ar}\left({\displaystyle \frac{39.95\text{g}}{\cancel{\text{mol}}\text{Ar}}}\right)=0.037\text{g Ar}$

The result is in agreement with our expectations as noted above, around 0.04 g $\mathrm{Ar}$.

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Example 2.5.3: Deriving Number of Atoms from Mass for an Element

Copper is commonly used to fabricate electrical wire (Figure 2.5.4). How many copper atoms are in 5.00 g of copper wire?

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The number of $\mathrm{Cu}$ atoms in the wire may be conveniently derived from its mass by a two-step computation: first calculating the molar amount of $\mathrm{Cu}$, and then using Avogadro’s number (N_A) to convert this molar amount to number of $\mathrm{Cu}$ atoms. Considering that the provided sample mass (5.00 g) is a little less than one-tenth the mass of 1 mole of $\mathrm{Cu}$ (~64 g), a reasonable estimate for the number of atoms in the sample would be on the order of one-tenth N_A, or approximately 10²² $\mathrm{Cu}$ atoms. Carrying out the two-step computation yields:

$5.00\cancel{\text{g}}\mathrm{Cu}\left({\displaystyle \frac{\cancel{\text{mol}}\mathrm{Cu}}{63.55\cancel{\text{g}}}}\right)\left({\displaystyle \frac{6.022\times {10}^{23}\text{atoms}}{\cancel{\text{mol}}}}\right)=4.74\times {10}^{22}\text{atoms of copper}$

The factor-label method yields the desired cancellation of units, and the computed result is on the order of 10²² as expected.

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Glossary

Avogadro’s number (N_A): experimentally determined value of the number of entities comprising 1 mole of substance, equal to 6.022 × 10²³ mol^-1

Atomic mass: mass in grams of 1 mole of a substance

mole: amount of substance containing the same number of atoms, molecules, ions, or other entities as the number of atoms in exactly 12 grams of ${\phantom{A}}_{\phantom{}}^{\phantom{12}}{\phantom{A}}_{\phantom{2}}^{\phantom{2}12}\mathrm{C}$